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Answer:</u></h3>
The reaction would progress faster and the activation energy would be lowered when the enzyme gets added.
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Explanation:</u></h3>
Enzymes are proteins that basically speed up the chemical reaction without being used. Enzymes are usually specific for a particular substrate. The substrate in the reaction bind to the active site of the enzyme which is present on the surface of the enzymes forming the enzyme-substrate complex.
Performing the enzyme-substrate complex the enzyme changes the shape slightly so that the substrate can fit tightly to its active site. Then this enzyme-substrate Complex undergoes a reaction to form a product. Enzymes lower the activation energy of a reaction i.e the required amount of energy needed for a reaction to occur.They do this by binding to a substrate and holding it in a way that allows the reaction to happen more efficiently.
Answer : The equilibrium concentration of 
will be, (C) 
Explanation : Given,
Equilibrium constant = 14.5
Concentration of 
at equilibrium = 0.15 M
Concentration of 
at equilibrium = 0.36 M
The balanced equilibrium reaction is,
The expression of equilibrium constant for the reaction will be:
![K_c=\frac{[CH_3OH]}{[CO][H_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3OH%5D%7D%7B%5BCO%5D%5BH_2%5D%5E2%7D)
Now put all the values in this expression, we get:
![14.5=\frac{[CH_3OH]}{(0.15)\times (0.36)^2}](https://tex.z-dn.net/?f=14.5%3D%5Cfrac%7B%5BCH_3OH%5D%7D%7B%280.15%29%5Ctimes%20%280.36%29%5E2%7D)
![[CH_3OH]=2.82\times 10^{-1}M](https://tex.z-dn.net/?f=%5BCH_3OH%5D%3D2.82%5Ctimes%2010%5E%7B-1%7DM)
Therefore, the equilibrium concentration of will be, (C)
The balanced chemical equation is Pb(NO3)2+ 2KI produces PbI2 + 2K(NO)3
.4 L of KI × (.375 mol/L of KI) × (1 mol of PbI2 / 2 mol of KI) × (461 g of PbI2/1 mol of PbI2)
=34. 6 g of PbI2 precipitate.
467.5 centimeters. are you sure that's high school chemistry though, lad?
In regards to this question, there are no options given to choose from and this makes the question difficult to answer. I hope the answer i am giving is the one you were looking for. The compound Mg(OH)2 when stirred in water will not pass through a filter paper as it is bound to form a sediment. This sediment will get stuck in the filter paper.
